Let $h(x)=x^5+x^4$. For what value of $x$ does $h$ have a relative maximum ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{5}{4}$ (Choice B) B $0$ (Choice C) C $-\dfrac{4}{5}$ (Choice D) D $1$
Solution: We can find the relative extrema (i.e. minima and maxima) of $h$ by looking for the intervals where its derivative $h'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $h$ is $h'(x)=x^3(5x+4)$. $h'(x)=0$ for $x=0,-\dfrac{4}{5}$. Since $h'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=0$ and $x=-\dfrac{4}{5}$. $h$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $x<-\frac{4}{5}$ $-\frac{4}{5}<x<0$ $-\frac{4}{5}$ $x>0$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $x<-\dfrac45$ $x=-1$ $h'(-1)=1>0$ $h$ is increasing $\nearrow$ $-\dfrac45<x<0$ $x=-\dfrac{1}{5}$ $h'\left(-\dfrac{1}{5}\right)=-\dfrac{3}{125}<0$ $h$ is decreasing $\searrow$ $x>0$ $x=1$ $h'(1)=9>0$ $h$ is increasing $\nearrow$ Now let's look at the critical points: $x$ Before After Verdict $-\dfrac45$ $\nearrow$ $\searrow$ Maximum $0$ $\searrow$ $\nearrow$ Minimum Now we can see that $h$ has a relative maximum at $x=-\dfrac{4}{5}$.